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(B) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \frac{\phi(y/x)}{\phi'(y/x)}$.Substitute $y = vx$, which implies $\frac{dy}{dx} =

Vedclass · Accepted Answer

$\phi\left(\frac{y}{x}\right) = kx$

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(B) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \frac{\phi(y/x)}{\phi'(y/x)}$.Substitute $y = vx$, which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting this into the equation gives: $v + x \frac{dv}{dx} = v + \frac{\phi(v)}{\phi'(v)}$.Subtracting $v$ from both sides, we get: $x \frac{dv}{dx} = \frac{\phi(v)}{\phi'(v)}$.Separating the variables, we have: $\frac{\phi'(v)}{\phi(v)} dv = \frac{dx}{x}$.Integrating both sides: $\int \frac{\phi'(v)}{\phi(v)} dv = \int \frac{dx}{x}$.This yields: $\ln|\phi(v)| = \ln|x| + \ln|k|$.Using logarithmic properties, $\phi(v) = kx$.Substituting $v = \frac{y}{x}$ back, we get: $\phi\left(\frac{y}{x}\right) = kx$.

The solution of the differential equation $\frac{dy}{dx} = \frac{y}{x} + \frac{\phi(y/x)}{\phi'(y/x)}$ is

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